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Current Question (ID: 7521)

Question:

$\text{If the size of the first orbit of the hydrogen atom is 5.29 nm, the size of the second orbit of He}^+ \text{ is}$

Options:

1. $2.65 \text{ nm}$
2. $21.16 \text{ nm}$
3. $10.58 \text{ nm}$
4. $5.29 \text{ nm}$

Solution:

$\text{Hint: Use hydrogen atom radius formula}$\n\n$\text{Step 1:}$\n\n$\text{The radius formula of a hydrogen atom is}$\n\n$r_n = \frac{a_0 n^2}{Z}$\n\n$\text{Here, } a_0 \text{ is the radius of hydrogen first orbital (Bohr radius), } Z \text{ is the atomic number, and } n \text{ is the principal quantum number of the orbit}$\n\n$\text{For the hydrogen atom in the first orbit, } n = 1 \text{ and } Z = 1\text{:}$\n\n$r_1(\text{H}) = \frac{a_0 \times 1^2}{1} = a_0 = 5.29 \text{ nm}$\n\n$\text{Step 2:}$\n\n$\text{The value of } Z \text{ for He}^+ \text{ is } 2 \text{ and the value of } n \text{ is } 2 \text{ for the second orbit.}$\n\n$\text{Put the value of } Z \text{ and } n \text{ in the above formula:}$\n\n$r_2(\text{He}^+) = \frac{a_0 \times 2^2}{2} = \frac{5.29 \times 4}{2} = 10.58 \text{ nm}$\n\n$\text{Therefore, the size of the second orbit of He}^+ \text{ is } 10.58 \text{ nm.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}